Essentially, i need this for a third year mathematics project and originally i thought i just needed to have something like this:
The group with this presentation is explicitly realized by the set of integers , under the operation of addition, where $a = 1$. Hence there exists an isomorphism from the infinite cyclic group of order $n$ to the integers mod $n$.
But i was told that i actually needed to show that there exists a free group that is generated by $a$, where $a$ is an element of the infinite cyclic group. Thank you.
A cyclic group by definition is one generated by one of its elements. This means that the set of elements $a^1, a^2, a^3, ...$ gives all elements of the group. For a finite group this implies that $a^n = 1$ (the identity element) for some n, n being the number of elements in the group (the group order).
For infinite cyclic groups the situation is similar, except that $a^0 = 1$ is added to the list, and $a^1, a^2, a^3, ...$ and$a^{-1}, a^{-2}, a^{-3}, ...$have to form the set of elements unequal to 1. So yes you need to find an element a that generates all group members. Note that $a^n$ may be read as $a+a+a+...+a$ (n times) for groups with additive + operation.